Why can protons and neutrons be close together in a nucleus despite the uncertainty principle?

Question

Vol II, Chapter 1 of the Feynman Lectures explains why negative electrons do not get closer to the positively charged nucleus despite the great attractive force:

If we try to confine our electrons in a region that is very close to the protons, then according to the uncertainty principle they must have some mean square momentum which is larger the more we try to confine them. It is this motion, required by the laws of quantum mechanics, that keeps the electrical attraction from bringing the charges any closer together.

Why can't this same argument be applied to the particles in the nucleus? I.e, why can we pinpoint the positions of adjacent protons and neutrons even though this would mean that their momentums are more uncertain and thus moving in such a way that they couldn't be co-residing in a nucleus?

Answer 1

Momentum is mass times velocity. Electrons have a teeny tiny mass (that's the technical term). Protons and neutrons are, relatively speaking, much much bigger (note, I'm not weight shaming any protons out there. I love how big you particles are <3).

Having more mass means they can be confined to a smaller volume without needing as large a velocity. They still have momenta that fits with the uncertainty principle.

Answer 2

The general rule for the size of a bound state is:

$$ R = \frac{\lambda_C}{2\pi\alpha} $$

where

$$\lambda_C = \frac{hc}{mc^2} = \frac{h}{mc}$$

is the Compton wavelength of the bound particle, and $\alpha$ is the dimensionless strength of the binding force.

Recall that if you confine a particle to less that its Compton wave length, the HUP gives it enough momentum (energy) to produce a matter/antimatter pair.

For the Hydrogen atom:

$$ R_{EM} = \frac{h}{2\pi m_e\alpha c} = 5.3\times 10^{-11}{\rm m} = a_0 $$

you get the Bohr radius.

For the proton, we have a larger mass (by a factor of 1835) and a stronger coupling ($\alpha_S \approx 1$), which is a factor of 137, so:

$$ R_{\rm Strong} = \frac{h}{2\pi M_p\alpha_S c} = 0.2\,{\rm fm}\approx R_p/5 $$

So the result is about 5 times too small. I attribute this to $\alpha_S$ being the quark coupling, not the effective hadronic field theory coupling (which is dimensional, see: https://cds.cern.ch/record/466436/files/0009312.pdf).

Perhaps an informed theorist can give a dimensionless $\alpha_{\pi}$?

Answer 3

Electrons in atoms have eV-scale kinetic and potential energies. Using the nonrelativistic kinetic energy

$$ T = \frac{p^2}{2m} = \frac{(pc)^2}{2mc^2} $$

we get typical momentum

$$ pc \sim \sqrt{T\cdot 2mc^2} \sim \sqrt{\text{eV}\cdot2\times\frac12\,\text{mega-eV}} \sim \text{kilo-eV} $$

corresponding to position uncertainty

$$ \sigma_x \sim \frac{\hbar c}{\sigma_p c} \sim \frac{197\,\rm eV\,nm}{\text{kilo-eV}} \sim \frac15\rm\,nm $$

This says that the valence electrons in an atoms should have orbitals a few angstroms across. Not bad for one-squiggle algebra. Now let's do the same thing for a nucleon, using a typical nuclear binding energy:

$$ p_\text{nucleon}c \sim \sqrt{T\cdot mc^2} \sim \sqrt{\text{mega-eV}\cdot\text{giga-eV}} \sim 30\rm\,MeV \\ \sigma_x c \sim\frac{197\rm\,MeV\,fm}{30\rm\,MeV} \sim \text{few}\rm\,fm $$

Nuclei are smaller than atoms because a bound nucleon has more momentum uncertainty than a bound electron.